Equivalences of skeletal categories are isomorphisms
$begingroup$
As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.
Does this hold?
category-theory
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
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add a comment |
$begingroup$
As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.
Does this hold?
category-theory
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
$endgroup$
add a comment |
$begingroup$
As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.
Does this hold?
category-theory
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
$endgroup$
As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.
Does this hold?
category-theory
category-theory
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
edited Dec 5 '18 at 16:21
Guido A.
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
asked Dec 5 '18 at 15:38
Guido A.Guido A.
7,3111730
7,3111730
add a comment |
add a comment |
1 Answer
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$begingroup$
It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.
A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.
So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
$endgroup$
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
add a comment |
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$begingroup$
It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.
A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.
So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
$endgroup$
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
add a comment |
$begingroup$
It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.
A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.
So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
$endgroup$
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
add a comment |
$begingroup$
It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.
A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.
So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
$endgroup$
It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.
A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.
So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
answered Dec 5 '18 at 18:18
Kevin CarlsonKevin Carlson
32.6k23271
32.6k23271
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
add a comment |
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
$begingroup$
Got it, thanks a lot!
$endgroup$
– Guido A.
Dec 5 '18 at 18:54
add a comment |
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